Yes both, as if both players have chains of same length then it is a win for player to play (A) (exception if there are just 5 free cells in common) , hence I gave the possibility for B to have a long-chain of length one more than A.

The pawns or structures existing in the free region is not taken into account.

5 free cells in common always allows B to win when they have same number of free cells and B long chain is one more than A.

Otherwise, if the number of free cells for each is equal and odd and greater than 1, B wins when the number of common free cells is 4 or 5.